Beton II Bab IV - 26

4.3.3 Tugas

Mutu Bahan : Fc’ = (30 + NO/2) Mpa

Fy = 240 Mpa atau 400 MPa

`

Gambar 4.13 Proyek Gorong gorong Jakarta 2011

Gambar 4.12 Rencana Gorong

Gorong gorong spt pada gambar. Tebal dinding adalah 15 cm

2m

0,5m

2m

Rencanakan dan Gambar Penulangan.

GORONG GORONG

TULANGAN yang TERSEDIA As(mm2)

1 250 200 150 125 100 Fy(Mpa)

D13 133 531 664 885 1062 1328 400 BJTD

D16 201 804 1005 1340 1608 2010 400 BJTD

D19 285 1140 1425 1900 2280 2850 400 BJTD

D22 380 1520 1900 2533 3040 3800 400 BJTD

KETDIA

jarak pusat ke pusat ( mm )

Beton II Bab IV - 27

4.3.4 Evaluasi

Gambar 4.14 Contoh penyelesaian Gorong gorong

25 kips

11,5

kN/m2 fc' =  30 Mpafy = 400 Mpatebal = 150 mm

48,5kN/m2

112,5 kN

P = 112,5 kN 11,25 tonPw = 25 kip

Beban berfaktor

1 L =Pw/E25/(4+0,06*S) 15,5105I = +20% = 18,6126 k/ft 9,1597

p = 276,986 kN/m

Span = 1500 mm 1,5 m 470,880,4921 ft

0,5 ft

Beton II Bab IV - 28

Gambar 4.15 Distribusi Momen Gorong gorong

3,389832 tanah ,5m ,5*2000*1,5 8,47458

15 kN/m2lantai atas ,25*24*1,5

9 kN/m2W roof 24 kN/m2

lantai dasar =25/(7*2000/3048) * 2,55,44286 13,6071 ksf

LL 1 roda 6643,59 kN/m2

DL dinding 2x,25x24x1,5/(lebar)18 kN/m2

lantai atas 24 kN/m2W bawah 42 kN/m2

tekanan tanah

pt = =(,5+,25/2)x20x1,518,75 kN/m

Δp = =(2,5-,25/2)x20x1,571,25 kN/m

276,9861 kN

24 kN/m

18,75 kN/m

71,25

Δq = 52,5

42 kN/m

A

C D

B

1/30 ΔqL ² + 

+1/12wL2

1/8PL+1/12wL2

1/20 ΔqL ² + 

1/16 ΔqL ² 

1/4PL  + 1/8wL2

+ 1/8wL2

Beton II Bab IV - 29

Gambar 4.16 Contoh penyelesaian Gaya Dalam

M-AB 51,9349 4,5 = 56,4349 kNm 103,86977 6,75 110,62M-BA kNm

M-AC 3,9375 3,51563 = 7,45313 kNm 7,3828125 5,27344 12,6563M-CA 5,90625 3,51563 = 9,42188 kNm kNm

M-AC 7,875 = 7,875 kNm 11,8125 11,8125M-CA kNm

1/8PL+1/12wL2 1/4PL  + 1/8wL2

1/30 ΔqL ² + 1/12wL2

1/20 ΔqL ² + /12wL2

1/16 ΔqL ² + 1/8wL2

+1/12wL2

‐7,453125 56,43489 ‐56,434886 7,453125‐24,49088 ‐24,49088 24,490881 24,49088052‐0,386719 12,24544 ‐12,24544 0,38671875‐5,929361 ‐5,929361 5,9293608 5,9293607552,96468 2,96468 ‐2,9646804 ‐2,964680377‐2,96468 ‐2,96468 2,9646804 2,9646803771,111755 1,48234 ‐1,4823402 ‐1,111755142‐1,297048 ‐1,297048 1,2970477 1,297047665

‐38,44538 38,44538 ‐38,445378 38,44537755‐38,44538 38,445378

9,421875 ‐7,875 7,875 ‐9,421875‐0,773438 ‐0,773438 0,7734375 0,7734375‐12,24544 0,386719 ‐0,3867188 12,245440265,929361 5,929361 ‐5,9293608 ‐5,929360755‐1,48234 ‐2,96468 2,9646804 1,4823401892,22351 2,22351 ‐2,2235103 ‐2,223510283

‐0,648524 ‐1,111755 1,1117551 0,6485238330,880139 0,880139 ‐0,8801395 ‐0,880139487

3,305144 ‐3,305144 3,3051437 ‐3,305143744

4 DESIGN As minM tumpuan 38,4454 kNm 630M Lapangan 72,1744 kNmSHEAR 161,028 kN

Vc = 73029,7 kN not OK0,0325

dbfc

Vc w ..6

'⎟⎟⎠

⎞⎜⎜⎝

⎛=  

0035,0400

4,14,1min ===

fyρ

OKxdb

As →>=== 0035,001309,045025062,1472

.ρ

600x0,85x30.6001c. βf'209960 penebalan

fc' = 30fy = 400tebal = 150penebala 300 0,0244

dbfc

Vc w ..6

'⎟⎟⎠

⎞⎜⎜⎝

⎛=  

( ) ( )

OK0,01309ρ0,0162565130,75x0,032b0,75x ρmakρ

,0325130400400600

600x 0,85x 300,85.

fyfy600

.6001c. βf'0,85bρ

→=>===

=+

=+

=

pmax min

As 4389,1875 630Mn = 268618275 43092000

214894620 34473600kNm 214,89462 34,4736

t = 250

0035,0400

4,14,1min ===

fyρ

OKxdb

As →>=== 0035,001309,045025062,1472

.ρ

( ) ( )

OK0,01309ρ0,0162565130,75x0,032b0,75x ρmakρ

,0325130400400600

600x 0,85x 300,85.

fyfy600

.6001c. βf'0,85bρ

→=>===

=+

=+

=

d = 180 tump lap pembagi0,215 As 702,583654 1318,9765 630

5 D!3 - 150 100 200

D13‐100 D13‐200 D13‐100

225,7395 kN

59,552 roda5 16 Wdl

D13‐200D13‐150 D13‐150

D13‐100D

5,16 Wdl0 M/L

64,712 jumlah

161,0275 kND13‐200

D13‐150 D13‐150

D13‐200D13‐10

D13‐100 D13‐200 D13‐100

00